As the options are not included with this question, we cannot chose a particular statement. Nevertheless, we are able to explain what the Supreme Court has ruled when it comes to the constitutional requirements of confinement.
The first case that addressed such conditions was that of <em>Holt v. Sarver</em>, in 1970. This was the first in a series of common law cases that found state prison systems to violate the Eighth Amendment. This amendment prohibits the use of cruel and unusual punishment.
This series of cases established that confinement should not include the unnecessary infliction of pain, nor should conditions be grossly disproportionate to the severity of the crime warranting imprisonment. The restrictions placed on prisoners can be restrictive, and even harsh, but should not become cruel. This includes the display of deliberate indifference in emergencies on the part of officials, or malicious and sadistic acts.
Answer:
an order to compel and impose sanctions for noncompliance
Explanation:
:)
Answer:
Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus. Any two circles can be made concentric by inversion by picking the inversion center as one of the limiting points.
1. Picking any two points on the outer circle and connecting them gives 1/3.
2. Picking any random point on a diagonal and then picking the chord that perpendicularly bisects it gives 1/2.
3. Picking any point on the large circle, drawing a line to the center, and then drawing the perpendicularly bisected chord gives 1/4.
So some care is obviously needed in specifying what is meant by "random" in this problem.
Given an arbitrary chord BB^' to the larger of two concentric circles centered on O, the distance between inner and outer intersections is equal on both sides (AB=A^'B^'). To prove this, take the perpendicular to BB^' passing through O and crossing at P. By symmetry, it must be true that PA and PA^' are equal. Similarly, PB and PB^' must be equal. Therefore, PB-PA=AB equals PB^'-PA^'=A^'B^'. Incidentally, this is also true for homeoids, but the proof is nontrivial.
